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6x^{4}-5x^{3}+7x^{2}-5x+1=0
To factor the expression, solve the equation where it equals to 0.
±\frac{1}{6},±\frac{1}{3},±\frac{1}{2},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.
x=\frac{1}{3}
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
2x^{3}-x^{2}+2x-1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{4}-5x^{3}+7x^{2}-5x+1 by 3\left(x-\frac{1}{3}\right)=3x-1 to get 2x^{3}-x^{2}+2x-1. To factor the result, solve the equation where it equals to 0.
±\frac{1}{2},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.
x=\frac{1}{2}
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
x^{2}+1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x^{2}+2x-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}+1. To factor the result, solve the equation where it equals to 0.
x=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 0 for b, and 1 for c in the quadratic formula.
x=\frac{0±\sqrt{-4}}{2}
Do the calculations.
x^{2}+1
Polynomial x^{2}+1 is not factored since it does not have any rational roots.
\left(2x-1\right)\left(3x-1\right)\left(x^{2}+1\right)
Rewrite the factored expression using the obtained roots.